5.6 Integrals involving exponential and logarithmic functions  (Page 4/4)

${\displaystyle \int e^{2x}}dx$

${\displaystyle \int e^{\mathrm{-3}x}}dx$

${\displaystyle \int 2^x}dx$

${\displaystyle \int 3^{\text{−}x}}dx$

${\displaystyle \int \frac{1}{2x}}dx$

${\displaystyle \int \frac{2}{x}}dx$

${\displaystyle \int \frac{1}{x^2}}dx$

${\displaystyle \int \frac{1}{\sqrt{x}}}dx$

${\displaystyle \int \frac{\text{ln}x}{x}}dx$

${\displaystyle \int \frac{dx}{x{\left(\text{ln}x\right)}^2}}$

${\displaystyle \int \frac{dx}{x\text{ln}x}}(x>1)$

${\displaystyle \int \frac{dx}{x\text{ln}x\text{ln}(\text{ln}x)}}$

${\displaystyle \int \text{tan}\theta d\theta }$

${\displaystyle \int \frac{\text{cos}x-x\text{sin}x}{x\text{cos}x}dx}$

${\displaystyle \int \frac{\text{ln}\left(\text{sin}x\right)}{\text{tan}x}dx}$

${\displaystyle \int \text{ln}(\text{cos}x)}\text{tan}xdx$

${\displaystyle \int x}{e}^{\text{−}{x}^2}dx$

${\displaystyle \int x^2}{e}^{\text{−}{x}^3}dx$

${\displaystyle \int e^{\text{sin}x}\text{cos}xdx}$

${\displaystyle \int e^{\text{tan}x}}{\text{sec}}^{2}xdx$

${\displaystyle \int e^{\text{ln}x}\frac{dx}{x}}$

${\displaystyle \int \frac{e^{\text{ln}(1-t)}}{1-t}}dt$

${\displaystyle \int \text{ln}xdx}$ $\text{(}Hint\text{:}{\displaystyle \int \text{ln}xdx=\frac{1}{2}{\displaystyle \int x}\text{ln}\left(x^2\right)dx}\text{)}$

${\displaystyle \int x^2}{\text{ln}}^{2}xdx$

${\displaystyle \int \frac{\text{ln}x}{x^2}}dx$ $\text{(}Hint\text{:}\text{Set}u=\frac{1}{x}\text{.}\text{)}$

${\displaystyle \int \frac{\text{ln}x}{\sqrt{x}}dx}$ $\text{(}Hint\text{:}\text{Set}u=\sqrt{x}\text{.}\text{)}$

Write an integral to express the area under the graph of $y=\frac{1}{t}$ from $t=1$ to e ^x and evaluate the integral.

Write an integral to express the area under the graph of $y=e^t$ between $t=0$ and $t=\text{ln}x,$ and evaluate the integral.

${\displaystyle \int \text{tan}\left(2x\right)dx}$

${\displaystyle \int \frac{\text{sin}\left(3x\right)-\text{cos}\left(3x\right)}{\text{sin}\left(3x\right)+\text{cos}\left(3x\right)}dx}$

${\displaystyle \int \frac{x\text{sin}\left(x^2\right)}{\text{cos}\left(x^2\right)}dx}$

${\displaystyle \int x\text{csc}\left(x^2\right)dx}$

${\displaystyle \int \text{ln}\left(\text{cos}x\right)\text{tan}xdx}$

${\displaystyle \int \text{ln}\left(\text{csc}x\right)\text{cot}xdx}$

${\displaystyle \int \frac{e^x-e^{\text{−}x}}{e^x+e^{\text{−}x}}dx}$

${\displaystyle {\int }_1^2\frac{1+2x+x^2}{3x+3x^2+x^3}dx}$

${\displaystyle {\int }_0^{\pi \text{/}4}\text{tan}xdx}$

${\displaystyle {\int }_0^{\pi \text{/}3}\frac{\text{sin}x-\text{cos}x}{\text{sin}x+\text{cos}x}dx}$

${\displaystyle {\int }_{\pi \text{/}6}^{\pi \text{/}2}\text{csc}xdx}$

${\displaystyle {\int }_{\pi \text{/}4}^{\pi \text{/}3}\text{cot}xdx}$

${\displaystyle \int \frac{x}{x-100}dx};u=x-100$

${\displaystyle \int \frac{y-1}{y+1}dy};u=y+1$

${\displaystyle \int \frac{1-x^2}{3x-x^3}dx};u=3x-x^3$

${\displaystyle \int \frac{\text{sin}x+\text{cos}x}{\text{sin}x-\text{cos}x}dx};u=\text{sin}x-\text{cos}x$

${\displaystyle \int e^{2x}}\sqrt{1-e^{2x}}dx;u=e^{2x}$

${\displaystyle \int \text{ln}\left(x\right)\frac{\sqrt{1-{\left(\text{ln}x\right)}^2}}{x}dx};u=\text{ln}x$

[T] $y=e^x$ over $\left[0,1\right]$

[T] $y=e^{\text{−}x}$ over $\left[0,1\right]$

[T] $y=\text{ln}\left(x\right)$ over $\left[1,2\right]$

[T] $y=\frac{x+1}{x^2+2x+6}$ over $\left[0,1\right]$

[T] $y=2^x$ over $\left[\mathrm{-1},0\right]$

[T] $y=\text{−}{2}^{\text{−}x}$ over $\left[0,1\right]$

$f\left(x\right)=\frac{{\text{log}}_{10}\left(x\right)}{x};a=10,b=100$

$f\left(x\right)=\frac{{\text{log}}_2\left(x\right)}{x};a=32,b=64$

$f\left(x\right)=2^{\text{−}x};a=1,b=2$

$f\left(x\right)=2^{\text{−}x};a=3,b=4$

Find the area under the graph of the function $f\left(x\right)=x{e}^{\text{−}{x}^2}$ between $x=0$ and $x=5.$

Compute the integral of $f\left(x\right)=x{e}^{\text{−}{x}^2}$ and find the smallest value of N such that the area under the graph $f\left(x\right)=x{e}^{\text{−}{x}^2}$ between $x=N$ and $x=N+10$ is, at most, 0.01.

Find the limit, as N tends to infinity, of the area under the graph of $f\left(x\right)=x{e}^{\text{−}{x}^2}$ between $x=0$ and $x=5.$

Show that ${\displaystyle {\int }_a^b\frac{dt}{t}}={\displaystyle {\int }_{1\text{/}b}^{1\text{/}a}\frac{dt}{t}}$ when $0<a\le b.$

Suppose that $f\left(x\right)>0$ for all x and that f and g are differentiable. Use the identity $f^g=e^{g\text{ln}f}$ and the chain rule to find the derivative of $f^g.$

Use the previous exercise to find the antiderivative of $h\left(x\right)=x^x\left(1+\text{ln}x\right)$ and evaluate ${\displaystyle {\int }_2^3{x}^x\left(1+\text{ln}x\right)dx}.$

Show that if $c>0,$ then the integral of $1\text{/}x$ from ac to bc $\left(0<a<b\right)$ is the same as the integral of $1\text{/}x$ from a to b .

Use the identity $\text{ln}\left(x\right)={\displaystyle {\int }_1^x\frac{dt}{t}}$ to derive the identity $\text{ln}\left(\frac{1}{x}\right)=\text{−}\text{ln}x.$

Use a change of variable in the integral ${\displaystyle {\int }_1^{xy}\frac{1}{t}dt}$ to show that $\text{ln}xy=\text{ln}x+\text{ln}y\text{for}x,y>0.$

Use the identity $\text{ln}x={\displaystyle {\int }_1^x\frac{dt}{x}}$ to show that $\text{ln}(x)$ is an increasing function of x on $[0,\infty ),$ and use the previous exercises to show that the range of $\text{ln}(x)$ is $\left(\text{−}\infty ,\infty \right).$ Without any further assumptions, conclude that $\text{ln}(x)$ has an inverse function defined on $\left(\text{−}\infty ,\infty \right).$

Pretend, for the moment, that we do not know that $e^x$ is the inverse function of $\text{ln}(x),$ but keep in mind that $\text{ln}(x)$ has an inverse function defined on $\left(\text{−}\infty ,\infty \right).$ Call it E . Use the identity $\text{ln}xy=\text{ln}x+\text{ln}y$ to deduce that $E\left(a+b\right)=E\left(a\right)E\left(b\right)$ for any real numbers a , b .

Pretend, for the moment, that we do not know that $e^x$ is the inverse function of $\text{ln}x,$ but keep in mind that $\text{ln}x$ has an inverse function defined on $\left(\text{−}\infty ,\infty \right).$ Call it E . Show that $E\text{'}\left(t\right)=E\left(t\right).$

The sine integral, defined as $S\left(x\right)={\displaystyle {\int }_0^x\frac{\text{sin}t}{t}dt}$ is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large x . Show that for $k\ge 1,\left|S\left(2\pi k\right)-S\left(2\pi \left(k+1\right)\right)\right|\le \frac{1}{k\left(2k+1\right)\pi }.$ $\text{(}Hint\text{:}\text{sin}\left(t+\pi \right)=\text{−}\text{sin}t\text{)}$

[T] The normal distribution in probability is given by $p\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\text{−}{\left(x-\mu \right)}^2\text{/}2{\sigma }^2},$ where σ is the standard deviation and μ is the average. The standard normal distribution in probability, $p_s,$ corresponds to $\mu =0\text{and}\sigma =1.$ Compute the left endpoint estimates $R_{10}\text{and}{R}_{100}$ of ${\displaystyle {\int }_{\mathrm{-1}}^1\frac{1}{\sqrt{2\pi }}{e}^{\text{−}{x}^{2\text{/}2}}dx}.$

[T] Compute the right endpoint estimates $R_{50}\text{and}{R}_{100}$ of ${\displaystyle {\int }_{\mathrm{-3}}^5\frac{1}{2\sqrt{2\pi }}{e}^{\text{−}{\left(x-1\right)}^2\text{/}8}.}$